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pentagon [3]
3 years ago
14

Which of the points shown below are on the line given by the equation y = 2/3 x? Check all that apply. ​

Mathematics
2 answers:
malfutka [58]3 years ago
3 0

Answer: The answers are A, C, D.

Hope this helps.

liberstina [14]3 years ago
3 0

The answer is A, C, and D

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Convert 1.4x10^-8 to standard notation/long form​
andriy [413]

Answer:

0.000000014

Step-by-step explanation:

1.4 × 10^{-8} = \frac{1.4}{10^8 = 100000000} = 0.000000014

6 0
2 years ago
Solve x^2+ 4x =15<br> thxssss
dmitriy555 [2]

Answer:

x=3/5

Step-by-step explanation:

x^2+4x=5

x+4x=5x

5x+2=5

-2 -2

5x=3

divide 5 from each side

5/5x=3/5

<u>x=3/5</u>

7 0
3 years ago
Given h(x)=-x-4, solve for x when h(x)=4
Brilliant_brown [7]

Step-by-step explanation:

h(x)=-x-4

but h(x)=4

4=-x-4

x=-4-4

x=-8

5 0
3 years ago
Write 3.08 × 10-5 as an ordinary number.
Vlad [161]

Answer:

0.0000308

Step-by-step explanation:

there are 5 places that it has moved from.

3 0
3 years ago
Each week, Heather’s company has $5000 in fixed costs plus an additional $250 for each system produced. The company is able to p
kvv77 [185]

The question is an illustration of composite functions.

  • Functions c(n) and h(n) are \mathbf{c(n) = 5000 + 250n} and \mathbf{n(h) = 5h}
  • The composite function c(n(h)) is \mathbf{c(n(h)) = 5000 + 1250h}
  • The value of c(n(100)) is \mathbf{c(n(100)) = 130000}
  • The interpretation is: <em>"the cost of working for 100 hours is $130000"</em>

The given parameters are:

  • $5000 in fixed costs plus an additional $250
  • 5 systems in one hour of production

<u>(a) Functions c(n) and n(h)</u>

Let the number of system be n, and h be the number of hours

So, the cost function (c(n)) is:

\mathbf{c(n) = Fixed + Additional \times n}

This gives

\mathbf{c(n) = 5000 + 250 \times n}

\mathbf{c(n) = 5000 + 250n}

The function for number of systems is:

\mathbf{n(h) = 5 \times h}

\mathbf{n(h) = 5h}

<u>(b) Function c(n(h))</u>

In (a), we have:

\mathbf{c(n) = 5000 + 250n}

\mathbf{n(h) = 5h}

Substitute n(h) for n in \mathbf{c(n) = 5000 + 250n}

\mathbf{c(n(h)) = 5000 + 250n(h)}

Substitute \mathbf{n(h) = 5h}

\mathbf{c(n(h)) = 5000 + 250 \times 5h}

\mathbf{c(n(h)) = 5000 + 1250h}

<u>(c) Find c(n(100))</u>

c(n(100)) means that h = 100.

So, we have:

\mathbf{c(n(100)) = 5000 + 1250 \times 100}

\mathbf{c(n(100)) = 5000 + 125000}

\mathbf{c(n(100)) = 130000}

<u>(d) Interpret (c)</u>

In (c), we have: \mathbf{c(n(100)) = 130000}

It means that:

The cost of working for 100 hours is $130000

Read more about composite functions at:

brainly.com/question/10830110

5 0
3 years ago
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